JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:(-1,1) \rightarrow(0, \pi)

be defined by

f(x)=\cot ^{-1} \frac{2 x}{1-x^2}

Then

View answer and explanation

Correct option: C

f

is both one-one and onto

Option C (f is both one-one and onto) is correct because the function is both injective (one-to-one) and surjective (onto) within the specified domain and range.

More Questions from FUNCTIONS

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

, where

f(x)=x(4-x)

, then

f^{-1}(x)

is given by: If

g(f(x))=|\sin x|

and

f(g(x))=(\sin \sqrt{x})^2

, then Let

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

be defined by

f\left( x \right) = x - \left[ x \right]

, then

{f^{ - 1}}\left( x \right)

equals Let

f: R \rightarrow R

be defined as

f(x)=x^4

, Choose the correct answer. Which of the following functions are odd functions Let

f(x)=(x+1)^2-1(x \geq-1)

. Then the set

S=\left\{x: f(x)=f^{-1}(x)\right\}

contains